Realistic mode question
- Thread starter Orion
- Start date
Lemniscate Biscuit
ㅤㅤHelp DeskㅤㅤRL10 Expert
Modder
Team Judge
TEAM HAWK
Moon Maker
Atlas
Under Pressure
Registered
MOTY 2023
Lemniscate Biscuit
ㅤㅤHelp DeskㅤㅤRL10 Expert
Modder
Team Judge
TEAM HAWK
Moon Maker
Atlas
Under Pressure
Registered
MOTY 2023
Altaïr
Space Stig, Master of gravity
Staff member
Head Moderator
Team Kolibri
Modder
TEAM HAWK
Atlas
Deja Vu
Under Pressure
Forum Legend
Aha, I feel like this is a question for me 
There's actually no need of the planets mass to calculate anything in the game. The reason for that is because in any formula, the planet mass (M) is ALWAYS multipled by G, the universal gravitational constant. For example, the local gravitational field is expressed that way:
g = G.M/r²
The same goes for the orbital period, which is obtained through the following relation:
T²/sma³ = 4.π²/(G.M)
For this reason, what is used in practice is the gravitational parameter:
μ = G.M
That way, the 2 formulas above can be rewritten that way:
g = μ/r²
T²/sma³ = 4.π²/μ
So now you may ask, how do I know the gravitational parameter right? Actually you already have the answer with this post: precisely with the formula g = μ/r². The planet file gives you the planet's radius, and the gravity at surface, so you can use this to deduce the gravitational parameter:
μ = g × radius²
Don't forget that in realistic radiuses are multiplied by 20, so for Earth you have g=9.8 m/s, and radius = 315,000 × 20 = 6,300,000.0 m.
With all this you normally have everything that's needed to calculate the orbital period.
There's actually no need of the planets mass to calculate anything in the game. The reason for that is because in any formula, the planet mass (M) is ALWAYS multipled by G, the universal gravitational constant. For example, the local gravitational field is expressed that way:
g = G.M/r²
The same goes for the orbital period, which is obtained through the following relation:
T²/sma³ = 4.π²/(G.M)
For this reason, what is used in practice is the gravitational parameter:
μ = G.M
That way, the 2 formulas above can be rewritten that way:
g = μ/r²
T²/sma³ = 4.π²/μ
So now you may ask, how do I know the gravitational parameter right? Actually you already have the answer with this post: precisely with the formula g = μ/r². The planet file gives you the planet's radius, and the gravity at surface, so you can use this to deduce the gravitational parameter:
μ = g × radius²
Don't forget that in realistic radiuses are multiplied by 20, so for Earth you have g=9.8 m/s, and radius = 315,000 × 20 = 6,300,000.0 m.
With all this you normally have everything that's needed to calculate the orbital period.
Orion
Nuclear bombs in space, die-hard WALL-E fan.
Modder
Team Judge
TEAM HAWK
Swingin' on a Star
Atlas
Fly me to the Moon
Under Pressure
Registered
Aha, I feel like this is a question for me
There's actually no need of the planets mass to calculate anything in the game. The reason for that is because in any formula, the planet mass (M) is ALWAYS multipled by G, the universal gravitational constant. For example, the local gravitational field is expressed that way:
g = G.M/r²
The same goes for the orbital period, which is obtained through the following relation:
T²/sma³ = 4.π²/(G.M)
For this reason, what is used in practice is the gravitational parameter:
μ = G.M
That way, the 2 formulas above can be rewritten that way:
g = μ/r²
T²/sma³ = 4.π²/μ
So now you may ask, how do I know the gravitational parameter right? Actually you already have the answer with this post: precisely with the formula g = μ/r². The planet file gives you the planet's radius, and the gravity at surface, so you can use this to deduce the gravitational parameter:
μ = g × radius²
Don't forget that in realistic radiuses are multiplied by 20, so for Earth you have g=9.8 m/s, and radius = 315,000 × 20 = 6,300,000.0 m.
With all this you normally have everything that's needed to calculate the orbital period.
There's actually no need of the planets mass to calculate anything in the game. The reason for that is because in any formula, the planet mass (M) is ALWAYS multipled by G, the universal gravitational constant. For example, the local gravitational field is expressed that way:
g = G.M/r²
The same goes for the orbital period, which is obtained through the following relation:
T²/sma³ = 4.π²/(G.M)
For this reason, what is used in practice is the gravitational parameter:
μ = G.M
That way, the 2 formulas above can be rewritten that way:
g = μ/r²
T²/sma³ = 4.π²/μ
So now you may ask, how do I know the gravitational parameter right? Actually you already have the answer with this post: precisely with the formula g = μ/r². The planet file gives you the planet's radius, and the gravity at surface, so you can use this to deduce the gravitational parameter:
μ = g × radius²
Don't forget that in realistic radiuses are multiplied by 20, so for Earth you have g=9.8 m/s, and radius = 315,000 × 20 = 6,300,000.0 m.
With all this you normally have everything that's needed to calculate the orbital period.
How did you get that formula for T? I've never seen it before.
Last edited:
Altaïr
Space Stig, Master of gravity
Staff member
Head Moderator
Team Kolibri
Modder
TEAM HAWK
Atlas
Deja Vu
Under Pressure
Forum Legend
That formula is easily obtained from a few basic formulas.
You have to use the areal velocity: it's the area swept out by the position vector per unit of time (see here). By the second Kepler's law, it's a constant. Newton's equations allow to establish that A = √(p.μ)/2
p is the orbital parameter of the orbit.
The area of an ellipse is calculated as "π×a×b", where a is the semi major axis, and b the semi minor axis.
With the usual orbital parameters, you have:
a = p/(1-e²)
b = p/√(1-e²)
The full area of the ellipse is swept out in a full period at the rate of A, the areal velocity, which means:
π×a×b = A×T
By substituting everything, you have:
π×p²/(1-e²)^(3/2) = √(p.μ)×T/2
Fractional exponents... Yummy!
So:
T = 2×π/√(μ) × (p/(1-e²))^(3/2)
p/(1-e²) is the semi-major axis (sma), so this is equivalent to:
T = 2×π/√(μ) × sma^(3/2)
By squaring the relation to get rid of the fractional exponent, you finally have the formula I gave:
T² = 4π²/μ × sma³
This is practically equivalent to the third Kepler's law that states that T²/sma³ is a constant. Kepler simply didn't precise the value of the constant.
You have to use the areal velocity: it's the area swept out by the position vector per unit of time (see here). By the second Kepler's law, it's a constant. Newton's equations allow to establish that A = √(p.μ)/2
p is the orbital parameter of the orbit.
The area of an ellipse is calculated as "π×a×b", where a is the semi major axis, and b the semi minor axis.
With the usual orbital parameters, you have:
a = p/(1-e²)
b = p/√(1-e²)
The full area of the ellipse is swept out in a full period at the rate of A, the areal velocity, which means:
π×a×b = A×T
By substituting everything, you have:
π×p²/(1-e²)^(3/2) = √(p.μ)×T/2
Fractional exponents... Yummy!
So:
T = 2×π/√(μ) × (p/(1-e²))^(3/2)
p/(1-e²) is the semi-major axis (sma), so this is equivalent to:
T = 2×π/√(μ) × sma^(3/2)
By squaring the relation to get rid of the fractional exponent, you finally have the formula I gave:
T² = 4π²/μ × sma³
This is practically equivalent to the third Kepler's law that states that T²/sma³ is a constant. Kepler simply didn't precise the value of the constant.
Altaïr
Space Stig, Master of gravity
Staff member
Head Moderator
Team Kolibri
Modder
TEAM HAWK
Atlas
Deja Vu
Under Pressure
Forum Legend
You forgot the square:
Gm is 9.8 × 6300000²
Also, don't forget to multiply this by sma^(3/2). Be careful that sma must be counted from the center of the planet, not from the surface! So if you want to calculate the period for a circular orbit of 100 km of altitude, your sma is actually 6400000 meters (planet radius + altitude).
In this case, the result is:
The result is in seconds, so in practice, it is 1 hour, 25 minutes and 58 seconds.
Gm is 9.8 × 6300000²
Also, don't forget to multiply this by sma^(3/2). Be careful that sma must be counted from the center of the planet, not from the surface! So if you want to calculate the period for a circular orbit of 100 km of altitude, your sma is actually 6400000 meters (planet radius + altitude).
In this case, the result is:
The result is in seconds, so in practice, it is 1 hour, 25 minutes and 58 seconds.